POJ 3522 Slim Span

Slim Span

Time Limit: 5000ms

Memory Limit: 65536KB

This problem will be judged on  PKU. Original ID: 3522
64-bit integer IO format: %lld      Java class name: Main

 

 

Given an undirected weighted graph G, you should find one of spanning trees specified as follows.

The graph G is an ordered pair (V, E), where V is a set of vertices {v1, v2, …, vn} and E is a set of undirected edges {e1, e2, …, em}. Each edge e ∈ E has its weight w(e).

A spanning tree T is a tree (a connected subgraph without cycles) which connects all the n vertices with n ? 1 edges. The slimness of a spanning tree T is defined as the difference between the largest weight and the smallest weight among the n ? 1 edges of T.

Figure 5: A graph  G and the weights of the edges

For example, a graph G in Figure 5(a) has four vertices {v1, v2, v3, v4} and five undirected edges {e1, e2, e3, e4, e5}. The weights of the edges are w(e1) = 3, w(e2) = 5, w(e3) = 6,w(e4) = 6, w(e5) = 7 as shown in Figure 5(b).

Figure 6: Examples of the spanning trees of  G

There are several spanning trees for G. Four of them are depicted in Figure 6(a)~(d). The spanning tree Ta in Figure 6(a) has three edges whose weights are 3, 6 and 7. The largest weight is 7 and the smallest weight is 3 so that the slimness of the tree Ta is 4. The slimnesses of spanning trees Tb, Tc and Td shown in Figure 6(b), (c) and (d) are 3, 2 and 1, respectively. You can easily see the slimness of any other spanning tree is greater than or equal to 1, thus the spanning tree Td in Figure 6(d) is one of the slimmest spanning trees whose slimness is 1.

Your job is to write a program that computes the smallest slimness.

 

Input

The input consists of multiple datasets, followed by a line containing two zeros separated by a space. Each dataset has the following format.

n	m
a1	b1	w1
	?
am	bm	wm

Every input item in a dataset is a non-negative integer. Items in a line are separated by a space. n is the number of the vertices and m the number of the edges. You can assume 2 ≤n ≤ 100 and 0 ≤ m ≤ n(n ? 1)/2. ak and bk (k = 1, …, m) are positive integers less than or equal to n, which represent the two vertices vak and vbk connected by the kth edge ek. wk is a positive integer less than or equal to 10000, which indicates the weight of ek. You can assume that the graph G = (V, E) is simple, that is, there are no self-loops (that connect the same vertex) nor parallel edges (that are two or more edges whose both ends are the same two vertices).

 

Output

For each dataset, if the graph has spanning trees, the smallest slimness among them should be printed. Otherwise, ?1 should be printed. An output should not contain extra characters.

 

Sample Input

4 5
1 2 3
1 3 5
1 4 6
2 4 6
3 4 7
4 6
1 2 10
1 3 100
1 4 90
2 3 20
2 4 80
3 4 40
2 1
1 2 1
3 0
3 1
1 2 1
3 3
1 2 2
2 3 5
1 3 6
5 10
1 2 110
1 3 120
1 4 130
1 5 120
2 3 110
2 4 120
2 5 130
3 4 120
3 5 110
4 5 120
5 10
1 2 9384
1 3 887
1 4 2778
1 5 6916
2 3 7794
2 4 8336
2 5 5387
3 4 493
3 5 6650
4 5 1422
5 8
1 2 1
2 3 100
3 4 100
4 5 100
1 5 50
2 5 50
3 5 50
4 1 150
0 0

Sample Output

1
20
0
-1
-1
1
0
1686
50

Source

Japan 2007

 

解题：求一颗生成树，最大边最小边的差最小。

 

枚举最小生成树。。。

 1 #include 
 2 #include 
 3 #include 
 4 #include 
 5 #include 
 6 #include 
 7 #include 
 8 #include 
 9 #include 
10 #include <string>
11 #include <set>
12 #include 
13 #define LL long long
14 #define pii pair
15 #define INF 0x3f3f3f3f
16 using namespace std;
17 const int maxn = 20000;
18 struct arc{
19     int u,v,w;
20 };
21 arc e[maxn];
22 int uf[maxn],n,m;
23 bool cmp(const arc &x,const arc &y){
24     return x.w < y.w;
25 }
26 int Find(int x){
27     if(x != uf[x]) uf[x] = Find(uf[x]);
28     return uf[x];
29 }
30 int kruskal(int p){
31     for(int i = 0; i < maxn; ++i) uf[i] = i;
32     int maxV = 0,cnt = 0;
33     for(int i = p; i < m; ++i){
34         int tx = Find(e[i].u);
35         int ty = Find(e[i].v);
36         if(tx != ty){
37             maxV = max(maxV,e[i].w);
38             cnt++;
39             uf[tx] = ty;
40         }
41     }
42     if(cnt == n-1) return maxV - e[p].w;
43     return INF;
44 }
45 int main() {
46     while(~scanf("%d %d",&n,&m)&&(n||m)){
47         for(int i = 0; i < m; ++i)
48             scanf("%d %d %d",&e[i].u,&e[i].v,&e[i].w);
49         sort(e,e+m,cmp);
50         int ans = INF;
51         for(int i = 0; i <= m - n + 1; ++i)
52             ans = min(ans,kruskal(i));
53         if(ans == INF) puts("-1");
54         else printf("%d\n",ans);
55     }
56     return 0;
57 }

View Code

 

转载于:https://www.cnblogs.com/crackpotisback/p/4095876.html